During a tennis match, a player serves the ball at 29.8 m/s, with the center of

Question

During a tennis match, a player serves the ball at 29.8 m/s, with the center of the ball leaving the racquet horizontally 2.49 m above the court surface. The net is 12.0 m away and0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number.

Explanation / Answer

horizontal acceleration=0

==> horizontal speed remains constant

veritcal acceleration=9.8 m/s^2, in downward direction

a) time taken to reach the net=horizontal distance/horizontal speed=12/29.8=0.4026 seconds

in that time, vertical distance covered, in donwward direction=0.5*9.8*0.4026^2=0.7945 m

hence at that time height from the ground=2.49-0.7945=1.6955 m

as the net reach 0.9 m hight, distance between center of the ball and top of the net=1.6955-0.9=0.7955 m

b)now, horizontal speed=29.8*cos(5)=29.686 m/s

time taken to reach the net=12/29.686=0.4042 seconds

initial vertical speed=29.8*sin(5)=2.5972 m/s

then distance travelled vertically=2.5972*0.4042+0.5*9.8*0.4042^2=1.85 m

hence height from the ground=2.49-1.85=0.64 m

height from the top of the net=0.64-0.9=-0.26 m

hence the ball does not clear the net.

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